Here is a simpler example that I hope convinces you that $V$ need not be continuous, even in the one dimensional case.
Take one dimensional Brownian motion $(B_t)$ and define the stopping time $\tau(\omega)=1_{(B_0(\omega)<0)}$. Then, for any bounded measurable $f$, we have $$V(x)=E_x[f(B_1)]1_{(-\infty,0)}(x)+f(x) 1_{[0,\infty)}(x).$$
The function $V$ can be made discontinuous at zero by choosing $f$ to have a strict maximum at $x=0$, since then $E_x[f(B_1)]$ E_x[f(B_1)] < f(0)$. Comment: You really cannot expect the function$V$to be continuous in general. The values of a typical stopping time$\tau$are intimately tied up with the sample paths of the Brownian motion; in your words$\tau$is "strongly correlated'' with$\omega$. It's in the definition of stopping time. The only stopping times that are independent of the Brownian motion are the deterministic ones. 1 Here is a simpler example that I hope convinces you that$V$need not be continuous, even in the one dimensional case. Take one dimensional Brownian motion$(B_t)$and define the stopping time$\tau(\omega)=1_{(B_0(\omega)<0)}$. Then, for any bounded measurable$f$, we have $$V(x)=E_x[f(B_1)]1_{(-\infty,0)}(x)+f(x) 1_{[0,\infty)}(x).$$ The function$V$can be made discontinuous at zero by choosing$f$to have a strict maximum at$x=0$, since then$E_x[f(B_1)]$Comment: You really cannot expect the function$V$to be continuous in general. The values of a typical stopping time$\tau$are intimately tied up with the sample paths of the Brownian motion; in your words$\tau$is "strongly correlated'' with$\omega\$. It's in the definition of stopping time.